Problem: Nauuo is a girl who loves playing chess.

One day she invented a game by herself which needs n chess pieces to play on a m× m chessboard. The rows and columns are numbered from 1 to m. We denote a cell on the intersection of the r-th row and c-th column as (r,c).

The game's goal is to place n chess pieces numbered from 1 to n on the chessboard, the i-th piece lies on (r_i,\,c_i), while the following rule is satisfied: for all pairs of pieces i and j, |r_i-r_j|+|c_i-c_j|≥|i-j|. Here |x| means the absolute value of x.

However, Nauuo discovered that sometimes she couldn't find a solution because the chessboard was too small.

She wants to find the smallest chessboard on which she can put n pieces according to the rules.

She also wonders how to place the pieces on such a chessboard. Can you help her?

Input

The only line contains a single integer n (1≤ n≤ 1000) — the number of chess pieces for the game.

Output

The first line contains a single integer — the minimum value of m, where m is the length of sides of the suitable chessboard.

The i-th of the next n lines contains two integers r_i and c_i (1≤ r_i,c_i≤ m) — the coordinates of the i-th chess piece.

If there are multiple answers, print any.

Examples

Input


2


Output


2
1 1
1 2

Input


4


Output


3
1 1
1 3
3 1
3 3

Note

In the first example, you can't place the two pieces on a 1×1 chessboard without breaking the rule. But you can place two pieces on a 2×2 chessboard like this:

<image>

In the second example, you can't place four pieces on a 2×2 chessboard without breaking the rule. For example, if you place the pieces like this:

<image>

then |r_1-r_3|+|c_1-c_3|=|1-2|+|1-1|=1, |1-3|=2, 1<2; and |r_1-r_4|+|c_1-c_4|=|1-2|+|1-2|=2, |1-4|=3, 2<3. It doesn't satisfy the rule.

However, on a 3×3 chessboard, you can place four pieces like this:

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int in(int &a, int &b, int &c, int &d) {
  return scanf("%d%d%d%d", &a, &b, &c, &d);
}
int in(int &a, int &b, int &c) { return scanf("%d%d%d", &a, &b, &c); }
int in(int &a, int &b) { return scanf("%d%d", &a, &b); }
int in(long long &a) { return scanf("%lld", &a); }
int in(int &a) { return scanf("%d", &a); }
int in(char *s) { return scanf("%s", s); }
int in(char &s) { return scanf("%c", &s); }
int in(long double &a) { return scanf("%Lf", &a); }
void out(int a) { printf("%d", a); }
void outln(int a) { printf("%d\n", a); }
void out(long long a) { printf("%lld", a); }
void outln(long long a) { printf("%lld\n", a); }
const long double pi = acos((long double)-1);
const long long inf = 0x3f3f3f3f;
const long double eps = 1e-8;
const int N = 2.1e5;
const long long mod = 1e9 + 7;
int sign(long double a) { return a < -eps ? -1 : a > eps; }
int db_cmp(long double a, long double b) { return sign(a - b); }
pair<int, int> ans[1100];
int main() {
  int n;
  in(n);
  int k = n / 2 + 1;
  int x1 = 1, y1 = 0, x2 = k, y2 = k + 1;
  outln(k);
  int l = 0, r = n + 1;
  int fla = 2;
  while (r - l > 1) {
    if (fla == 0) {
      x1++;
      l++;
      ans[l] = pair<int, int>(x1, y1);
    } else if (fla == 1) {
      x2--;
      r--;
      ans[r] = pair<int, int>(x2, y2);
    } else if (fla == 2) {
      y1++;
      l++;
      ans[l] = pair<int, int>(x1, y1);
    } else {
      y2--;
      r--;
      ans[r] = pair<int, int>(x2, y2);
    }
    fla = (fla + 1) % 4;
  }
  for (int i = 1; i <= n; i++) {
    printf("%d %d\n", ans[i].first, ans[i].second);
  }
  return 0;
}